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igomit [66]
3 years ago
10

Alden, a passenger on a yacht moored 15 miles due north of a straight, east-west

Physics
1 answer:
Ghella [55]3 years ago
6 0

Answer:

To minimize the travel time from the yacht to the hospital, the motorboat should head in a direction of 12.83 degrees west of south.

Explanation:

If we assume that both the motorboat and ambulance will be moving at a constant speed, we can calculate the time that each one will take to travel a given distance using the following equation:

time=\frac{distance}{speed}

Then the total travel time from the yacht to the hospital will be the motorboat travel time plus the ambulance travel time

t=t_m+t_a

t=\frac{d_m}{s_m} +\frac{d_a}{s_a}

First we must write the total travel time in terms of the motorboat's direction (Θ).

cos(\theta)=\frac{15}{d_m}

d_m=\frac{15}{cos(\theta)}=15 sec(\theta)

d_a=60-d_1

tan(\theta)=\frac{d_1}{15}

d_1=15tan(\theta)

d_a=60-15tan(\theta)

t=t_m+t_a

t=\frac{d_m}{s_m} +\frac{d_a}{s_a}

t=\frac{15sec(\theta)}{20} + \frac{60-15tan(\theta)}{90}

t=\frac{15}{20}sec(\theta) + \frac{60}{90}-\frac{15}{90}tan(\theta)

t=\frac{3}{4}sec(\theta)-\frac{1}{6}tan(\theta) + \frac{2}{3}

 

So this last equation represents the variation of the total travel time as a function of the motorboat's direction.

To find the equation's minimum point (which would be the direction with the minimum total travel time), we must find \frac{dt}{d\theta} and then find its roots (its x-interceptions).

\frac{dt}{d\theta}=\frac{d}{d\theta} (\frac{3}{4}sec(\theta)-\frac{1}{6}tan(\theta) + \frac{2}{3})

\frac{dt}{d\theta}=\frac{3}{4}sec(\theta)tan(\theta)-\frac{1}{6}sec^2(\theta)

\frac{dt}{d\theta}=sec(\theta)(\frac{3}{4}tan(\theta)-\frac{1}{6}sec(\theta))

Now let's find the values of x which make \frac{dt}{d\theta}=0

\frac{dt}{d\theta}=sec(\theta)(\frac{3}{4}tan(\theta)-\frac{1}{6}sec(\theta))=0

As sec(\theta) is never equal to zero, then \frac{dt}{d\theta} would be zero when

\frac{3}{4}tan(\theta)=\frac{1}{6}sec(\theta)

Graphing both equations we can find their interceptions and this would the value we're looking for.

In the attached images we can see that \theta=0.224 rad=12.83° is the minimum point for t(\theta). Then, to minimize the travel time from the yacht to the hospital, the motorboat should head in a direction of 12.83 degrees west of south.

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a. I've attached a plot of the surface. Each face is parameterized by

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\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k

Then integrate the dot product of <em>f</em> with each normal vector over the corresponding face.

\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx

=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0

\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du

\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8

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c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since <em>S</em> is closed, we can find the total flux by applying the divergence theorem.

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where <em>R</em> is the interior of <em>S</em>. We have

\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7

The integral is easily computed in cylindrical coordinates:

\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2

\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3

as expected.

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Explanation:

The weight of the astronaut is given by the equation

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m is the mass of the astronaut

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g=\frac{GM}{r^2}

where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as

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When the astronaut is on the Earth's surface, r=R (where R is the Earth's radius), so his weight is

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Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

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Learn more about gravitational force:

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